Answer
\[\frac{x(2x+3)}{(x+1)(x^2+2x+2)}=\frac{-1}{(x+1)}+\frac{3x+2}{x^2+2x+2}\]
Work Step by Step
Because $x^2+2x+2$ has no real zero so we can not factor it so
In this case the general form of the partial fraction decomposition is
\[\frac{x(2x+3)}{(x+1)(x^2+2x+2)}=\frac{A}{(x+1)}+\frac{Bx+C}{x^2+2x+2}\;\;\;...(*)\]
\[2x^2+3x=A(x^2+2x+2)+(Bx+C)(x+1)\]
\[2x^2+3x=A(x^2+2x+2)+(Bx^2+Bx+Cx+C)\]
\[2x^2+3x=(A+B)x^2+(2A+B+C)x+(2A+C)\]
Compare like coefficients both side
\[A+B=2\;\;\;...(1)\]
\[2A+B+C=3\;\;\;...(2)\]
\[2A+C=0\Rightarrow C=-2A\;\;\;...(3)\]
Using (3) in (2) then (2) becomes
\[B=3\]
From (1)
\[A=-1\]
From (3)
\[C=2\]
From (*)
\[\frac{x(2x+3)}{(x+1)(x^2+2x+2)}=\frac{-1}{(x+1)}+\frac{3x+2}{x^2+2x+2}\]
Hence,
\[\frac{x(2x+3)}{(x+1)(x^2+2x+2)}=\frac{-1}{(x+1)}+\frac{3x+2}{x^2+2x+2}.\]
